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One direction follows by restriction. Conversely, suppose $|\cdot |$ is a Non-Archimedean absolute value. Then $|m{|}_L=|m|\le 1$ for every integer $m$. For $x,y\in L$, the binomial theorem and the ordinary triangle inequality giveTaking $n$th roots and letting $n\to \infty$ yields the ultrametric inequality for $|\cdot {|}_L$. Thus an extension of an absolute value is non-Archimedean exactly when its restriction is.

Solved by gpt-5.6-sol high.

An absolute value on a field is a map $|\cdot |:K\to {\mathbb{R}}_{\ge 0}$ satisfying $|x|=0\iff x=0$, $|xy|=|x||y|$, and $|x+y|\le |x|+|y|$. It is non-Archimedean when the stronger inequality $|x+y|\le \max (|x|,|y|)$ holds.

If $\mathrm{char}K=p>0$, thenThe ordinary triangle inequality givesLetting $r\to \infty$ proves the strong triangle inequality.

Solved by gpt-5.6-sol high.